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4 years ago

Light is traveling in air at a speed of 300,000 km/s. Which substance would cause the light to refract the LEAST?

1 answer:

6 0

Use the formula n = c / v, where c is the speed of light and v is the speed of light in the environment, that is, use the speed in brackets and the correct answer will be the one that will be the smallest result

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These types of electromagnetic waves are listed in order of increasing frequency:

xenn [34]

Answer:

B. X-rays

Explanation:

From the given choices, x-rays will have the highest energy of the given waves.

The energy of electromagnetic waves is highly dependent on their frequency and wavelength.

Electromagnetic waves with a high frequency and small wavelength will have higher energy compared to those with low frequency and high wavelength.

X-rays are one the most energetic waves on the periodic table. They have a very high frequency and low wavelength.

7 0

3 years ago

Read 2 more answers

The potential energy of a 1.7 x 10-3 kg particle is described by U (x )space equals space minus (17 space J )space cos open squa

hichkok12 [17]

Answer:

Explanation:

Given a particle of mass

M = 1.7 × 10^-3 kg

Given a potential as a function of x

U(x) = -17 J Cos[x/0.35 m]

U(x) = -17 Cos(x/0.35)

Angular frequency at x = 0

Let find the force at x = 0

F = dU/dx

F = -17 × -Sin(x/0.35) / 0.35

F = 48.57 Sin(x/0.35)

At x = 0

Sin(0) =0

Then,

F = 0 N

So, from hooke's law

F = -kx

Then,

0 = -kx

This shows that k = 0

Then, angular frequency can be calculated using

ω = √(k/m)

So, since k = 0 at x = 0

Then,

ω = √0/m

ω = √0

ω = 0 rad/s

So, the angular frequency is 0 rad/s

4 0

3 years ago

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance

maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = $X_{L} = j\omega L = 2\pi fL$

where

R = resistance

$X_{L} = Inductive Reactance$

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

$C = \frac{\epsilon_{o}A}{x}$

where

x = separation between the parallel plates

Thus

C ∝ $\frac{1}{x}$

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

Also,

Z ∝ I

Therefore,

$\frac{Z}{I} = \frac{Z'}{I'}$

$\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}$

${R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})$

${R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})$

Solving the above eqn:

$X_{C} = 4R$

6 0

3 years ago

Citrus2011 [14]

Answer:

net force is positive downward..B

6 0

3 years ago

-g A small block is attached to an ideal spring and is moving in SHM on a horizontal frictionless surface. The amplitude of the

Nina [5.8K]

Answer:

a= 92. 13 m/s²

Explanation:

Given that

Amplitude ,A= 0.165 m

The maximum speed ,V(max) = 3.9 m/s

We know that maximum velocity in the SHM given as

V(max) = ω A

ω=Angular speed

A=Amplitude

$\omega =\dfrac{3.9}{0.165}\ rad/s$

ω=23.63 rad/s

The maximum acceleration given as

a = ω² A

a= (23.63)² x 0.165 m/s²

a= 92. 13 m/s²

Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².

5 0

3 years ago